\(I\in d:\left\{{}\begin{matrix}x=-1+2t\\y=2t\\z=-4+t\end{matrix}\right.\left(t\in Z\right)\)
\(\Rightarrow I\left(-1+2t;2t;-4+t\right)\) và \(M\left(4;5;1\right)\)
\(\Rightarrow\overrightarrow{IM}=\left(5-2t;5-2t;5-t\right)\)
\(\Rightarrow R^2=IM^2=\left(5-2t\right)^2+\left(5-2t\right)^2+\left(5-t\right)^2\)
\(d\left(I;\left(P\right)\right)=\dfrac{\left|2\cdot\left(-1+2t\right)+2\cdot2t-\left(-4+t\right)\right|}{\sqrt{2^2+2^2+\left(-1\right)^2}}=\dfrac{\left|7t+2\right|}{3}\)
\(\Rightarrow d^2\left(I;\left(P\right)\right)=\dfrac{\left(7t+2\right)^2}{9}\)
\(R^2=d^2\left(I;\left(P\right)\right)+r^2\)
\(\Rightarrow\left(5-2t\right)^2+\left(5-2t\right)^2+\left(5-t\right)^2=\dfrac{\left(7t+2\right)^2}{9}+25\)
\(\Leftrightarrow16t^2-239t+223=0\Leftrightarrow\left[{}\begin{matrix}t=1\\t=\dfrac{223}{16}\left(l\right)\end{matrix}\right.\)
\(\Rightarrow I\left(1;2;-3\right)\Rightarrow\left\{{}\begin{matrix}a=1\\b=2\\c=-3\end{matrix}\right.\)
Vậy \(a+b+c=0\)
Chọn B.
